alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

The sign is due to negative fx negative direction

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Unformatted text preview: Notice that y ’(t) changes SIGN at T/2 . So we may break up the integral into 2 parts. t1 ∫ t4 y ’(t)dt = T/ 2 ∫ t4 y ’(t)dt + T/ 2 t1 ∫ T/ 2 y ’(t)dt t1 = [ y(t) ] + [ y(t) ] t4 T/ 2 = { y(T/2) -- y(t4) } + { y(t1) -- y(T/2)} = y(t1) -- y(t4) 145 From the Geometric point of view : Geometric t1 ∫ T/ 2 ∫ t4 So : ∫ t4 y ’(t)dt + t1 ∫ T/ 2 y ’(t)dt y ’(t)dt = (negative function) . (negative direction) = positive area : + A t1 ∫ T/ 2 t4 y ’(t)dt = T/ 2 y ’(t)dt = (positive function) . (negative direction) = negative area : − A t1 ∫ t4 y ’(t)dt = + A − A = 0 From the Analytical point of view : Analytical t1 ∫ t4 y ’(t)dt = y (t1) -- y(t4) = 0 There is NO CHANGE in the height y(t) over interval [ t1, t 4 ] . In general, let f(x) be a continuous function over the interval [a, b] and continuous contin F(x) = ANTIDERIVATIVE {f(x)} . Then : area under f(x) = 0 ov er the inter v al [ a, b] ⇔ CHANGE in F(x) = 0 ov interv over the inter v al [ a, b] We note that, even if f(x) changes sign several times over the interval [a, b] , we may still mechanically look up our Table of Integr als to find F(x) and directly Integ als compute the CHANGE. It is NOT necessary to break up the integration of f(x) into positive and negative areas as we did above. Let us see a few more examples. 146 Consider F (x) = x 2 . Then f(x) = F' (x) = 2 x. See the graph. continuous c ontin Example1: Integrate contin uous f (x) over the inter val [1, 2] going LEFT to RIGHT in the positive direction . p ositive Using the tr a pe z oidal method a n ar row strip or infinitesimal element infinitesimal t r pez inf of area is : [f(x i ) + f(x i -- 1) ] . δ x i where δ x i = x i -- x i −1 x f ( -- i ) . δ x i = 2 f(x i ) + f(x i -- 1) and -- i i s the MEAN POINT in [x i -- 1, x i ] s uch that f( -- i ) = x x . 2 This is the area of a narrow trapezium where the parallel sides are f( x i ) and f( x ) , and the perpendicular distance is δ x . We do not need to know -- . x i -- 1 i n →∞ F (x) = ∫ f (x)dx = LIMIT i=1 i Σ f ( -- i ) . δ x...
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